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Popular >

ay^'+y=c

Solution

a y^{^{'}} + y = c

Solution

y = - e^{\frac{- t - c _{1}}{a}} + c

Solution steps

a y^{'} (t) + y = c

Solve separable ODE:

y = - e^{\frac{- t - c _{1}}{a}} + c

y = - e^{\frac{- t - c _{1}}{a}} + c

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Frequently Asked Questions (FAQ)

What is the solution for ay^'+y=c ?
The solution for ay^'+y=c is y=-e^{(-t-c_{1})/a}+c